What is the extraneous solution to these equations? $\dfrac{x^2 - 8}{x - 9} = \dfrac{14x - 56}{x - 9}$
Multiply both sides by $x - 9$ $ \dfrac{x^2 - 8}{x - 9} (x - 9) = \dfrac{14x - 56}{x - 9} (x - 9)$ $ x^2 - 8 = 14x - 56$ Subtract $14x - 56$ from both sides: $ x^2 - 8 - (14x - 56) = 14x - 56 - (14x - 56)$ $ x^2 - 8 - 14x + 56 = 0$ $ x^2 + 48 - 14x = 0$ Factor the expression: $ (x - 8)(x - 6) = 0$ Therefore $x = 8$ or $x = 6$ The original expression is defined at $x = 8$ and $x = 6$, so there are no extraneous solutions.